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Interpolation and Implicit Regularization · Tutorial 947 of 989

Signal Alignment And Interpolator Risk

Compare interpolation risks for equal-norm signals aligned with different covariance eigendirections, and identify the geometric feature that determines the comparison.

Advanced 11 min read

What You'll Learn

  • Define conditional prediction risk for a noiseless, correctly specified linear model.
  • Express the minimum-norm interpolator’s coefficient error using the training design’s nullspace.
  • Compute directional risk in covariance eigen-coordinates.
  • Compare equal-norm signals in high- and low-variance directions using explicit designs.
  • Explain why covariance eigenvalues alone do not determine which direction has lower risk.

Why Signal Direction Matters

Two regression signals can have the same coefficient norm and still lead to different prediction risks after interpolation. The reason is that interpolation does not generally recover every component of the signal: the minimum-norm solution keeps the part visible to the training design and discards the part in its nullspace. Prediction risk then weights the resulting error according to the covariance of a new feature vector. Thus the signal’s direction interacts with both the training geometry and the test-feature covariance.

We study this interaction in a noiseless, correctly specified linear model. The comparison is conditional on a fixed training design, so it isolates signal direction from variation across possible training samples. It also makes an important qualification precise: “high-variance direction” by itself does not imply either higher or lower interpolation risk.

Model and Risk Definition

Definition: Let \(x\in\mathbb{R}^p\) be a random test feature vector with \(E[x]=0\) and covariance matrix \(\Sigma=E[xx^T]\), assumed positive definite. The noiseless response is \(y=x^T\beta^\ast\). For a coefficient vector \(\hat\beta\), its prediction risk is $$ R(\hat\beta;\beta^\ast)=E\left[(x^T\hat\beta-x^T\beta^\ast)^2\right] =(\hat\beta-\beta^\ast)^T\Sigma(\hat\beta-\beta^\ast). $$ The expectation is over a fresh test feature vector, with the training data held fixed.

The equality follows by writing the prediction error as \(x^T(\hat\beta-\beta^\ast)\) and using \(E[xx^T]=\Sigma\). Since the response has no additional noise, this risk is the squared prediction error relative to the true regression function. We do not include a separate irreducible-noise term.

Let the fixed training design be \(X\in\mathbb{R}^{n\times p}\), with full row rank and \(n<p\), and let \(y=X\beta^\ast\). Among all coefficient vectors that interpolate, the minimum-Euclidean-norm interpolator is $$ \hat\beta=X^T(XX^T)^{-1}y. $$ Full row rank ensures that \(XX^T\) is invertible. The row space of \(X\) is the subspace spanned by its rows; its orthogonal complement is the nullspace \(\mathcal{N}(X)=\{v:Xv=0\}\).

The Error Is the Signal’s Nullspace Component

Theorem: For a full-row-rank design \(X\), the minimum-norm interpolator satisfies $$ \hat\beta=P_{\operatorname{row}(X)}\beta^\ast,\qquad \hat\beta-\beta^\ast=-P_{\mathcal{N}(X)}\beta^\ast. $$ Consequently, its conditional prediction risk is $$ R(\hat\beta;\beta^\ast) =(P_{\mathcal{N}(X)}\beta^\ast)^T \Sigma(P_{\mathcal{N}(X)}\beta^\ast). $$

Proof. Substitute \(y=X\beta^\ast\) into the minimum-norm formula:

$$ \hat\beta=X^T(XX^T)^{-1}X\beta^\ast. $$

The matrix \(P=X^T(XX^T)^{-1}X\) is symmetric, because \(XX^T\) and its inverse are symmetric. It is idempotent:

$$ P^2=X^T(XX^T)^{-1}XX^T(XX^T)^{-1}X =X^T(XX^T)^{-1}X=P. $$

Its range is contained in the row space of \(X\), since \(Pv\) is a linear combination of the columns of \(X^T\). Conversely, if \(u=X^Ta\) is in that row space, then

$$ Pu=X^T(XX^T)^{-1}XX^Ta=X^Ta=u. $$

Therefore \(P\) is the orthogonal projector onto the row space. The orthogonal complement of that space is \(\mathcal{N}(X)\), so \(I-P=P_{\mathcal{N}(X)}\). Hence \(\hat\beta-\beta^\ast=-(I-P)\beta^\ast\). Substitution into the risk definition gives the stated quadratic form; the minus sign cancels. \(\square\)

This result separates the two ingredients of risk. The training design determines which part of the signal is missed, through \(P_{\mathcal{N}(X)}\). The covariance \(\Sigma\) determines how costly that missed component is for prediction. A coefficient error of a given Euclidean size can therefore have very different prediction costs in different directions.

Reading Directional Risk in Eigen-Coordinates

Suppose the covariance has orthonormal eigenvectors \(u_1,\ldots,u_p\) and eigenvalues \(\lambda_1,\ldots,\lambda_p>0\), so that \(\Sigma u_j=\lambda_j u_j\). A high-variance eigendirection has a larger eigenvalue; a low-variance eigendirection has a smaller one. Write \(q=P_{\mathcal{N}(X)}\beta^\ast\) for the coefficient error up to sign, and \(q_k=u_k^Tq\). The risk is

$$ R(\hat\beta;\beta^\ast)=q^T\Sigma q =\sum_{k=1}^p\lambda_k q_k^2. $$

In particular, if the signal is \(\beta^\ast=B u_j\), with norm \(|B|\), then

$$ R(\hat\beta;B u_j) =B^2\sum_{k=1}^p\lambda_k \left(u_k^TP_{\mathcal{N}(X)}u_j\right)^2. $$

The same signal norm \(B^2\) is multiplied by a direction-dependent quantity. The covariance eigenvalue \(\lambda_j\) matters, but it is not the only relevant eigenvalue: the projection of \(u_j\) into the training nullspace can have components in several covariance eigendirections.

Takeaway: Equal-norm signals have equal interpolation risk only when their nullspace projections have the same covariance-weighted squared length. Comparing their covariance eigenvalues alone is not enough.

Worked Comparisons

Worked Example: A Two-Dimensional Design

Let \(p=2\), \(n=1\), and take the one-row design \(X=(1,2)\). Let test features have covariance \(\Sigma=\operatorname{diag}(4,1)\), so the first coordinate is a high-variance eigendirection and the second is a low-variance eigendirection. Compare signals \(\beta_H^\ast=(B,0)^T\) and \(\beta_L^\ast=(0,B)^T\), which both have norm \(|B|\).

A vector spanning the nullspace of \(X\) is \(v=(-2,1)^T\), because \(Xv=-2+2=0\), and \(v^Tv=5\). Thus \(P_{\mathcal{N}(X)}=vv^T/5\). For the high-variance signal,

$$ P_{\mathcal{N}(X)}\beta_H^\ast =\frac{v(v^T\beta_H^\ast)}{5} =\frac{(-2,1)^T(-2B)}{5} =\left(\frac{4B}{5},-\frac{2B}{5}\right)^T. $$

Its risk is

$$ 4\left(\frac{4B}{5}\right)^2+ \left(-\frac{2B}{5}\right)^2 =\frac{64B^2}{25}+\frac{4B^2}{25} =\frac{68B^2}{25}. $$

For the low-variance signal, the nullspace projection is

$$ P_{\mathcal{N}(X)}\beta_L^\ast =\frac{v(v^T\beta_L^\ast)}{5} =\frac{(-2,1)^T B}{5} =\left(-\frac{2B}{5},\frac{B}{5}\right)^T, $$

and its risk is

$$ 4\left(-\frac{2B}{5}\right)^2+ \left(\frac{B}{5}\right)^2 =\frac{16B^2}{25}+\frac{B^2}{25} =\frac{17B^2}{25}. $$

The high-direction signal has four times the risk here: \((68/25)/(17/25)=4\). The design places a larger fraction of that signal in its nullspace, and the resulting error also has a component in the high-variance coordinate.

Worked Example: A Three-Dimensional Design Favors the High-Variance Signal

Let \(\Sigma=\operatorname{diag}(9,4,1)\). The first coordinate is the high-variance direction and the third is the low-variance direction. Take \(X\) to have rows \((2,-1,0)\) and \((3,0,-1)\). These rows are linearly independent, and both are orthogonal to \(v=(1,2,3)^T\). Since the nullspace is one-dimensional, it is exactly the span of \(v\).

The squared norm of \(v\) is \(1+4+9=14\), so the nullspace projector is \(vv^T/14\). Also,

$$ v^T\Sigma v=9(1)^2+4(2)^2+1(3)^2=9+16+9=34. $$

For \(\beta_H^\ast=B(1,0,0)^T\), the nullspace projection is \(Bv/14\). Therefore,

$$ R(\hat\beta;\beta_H^\ast) =\frac{B^2}{14^2}v^T\Sigma v =\frac{34B^2}{196} =\frac{17B^2}{98}. $$

For \(\beta_L^\ast=B(0,0,1)^T\), the projection is \(3Bv/14\), because the inner product of \(v\) with the third coordinate vector is \(3\). Thus,

$$ R(\hat\beta;\beta_L^\ast) =\frac{9B^2}{14^2}v^T\Sigma v =\frac{306B^2}{196} =\frac{153B^2}{98}. $$

Both signals have norm \(|B|\), yet the low-direction signal has nine times the risk. In this design, it has a larger projection onto the nullspace. This example shows why the word “high-variance” does not, on its own, identify the signal with greater risk.

Worked Example: Reversing the Risk Ordering

Keep \(\Sigma=\operatorname{diag}(9,4,1)\), but use rows \((1,-3,0)\) and \((1,0,-3)\) for \(X\). They are linearly independent and orthogonal to \(v=(3,1,1)^T\), so \(v\) spans the nullspace. Here \(v^Tv=9+1+1=11\), and

$$ v^T\Sigma v=9(3)^2+4(1)^2+1(1)^2=81+4+1=86. $$

For the high-direction signal \(B(1,0,0)^T\), its inner product with \(v\) is \(3B\); for the low-direction signal \(B(0,0,1)^T\), the inner product is \(B\). The squared nullspace projection factors are therefore \(9B^2/11^2\) and \(B^2/11^2\), respectively. Their risks are

$$ R_H=\frac{9B^2}{121}(86)=\frac{774B^2}{121}, \qquad R_L=\frac{B^2}{121}(86)=\frac{86B^2}{121}. $$

Now the high-direction signal has nine times the risk. Both this example and the preceding one use the same covariance and equal-norm signals; changing the training geometry reverses the comparison.

A Useful Special Case and a Common Pitfall

When the training nullspace is one-dimensional, the comparison has a compact form. If it is spanned by a nonzero vector \(v\), then \(P_{\mathcal{N}(X)}=vv^T/(v^Tv)\). For a signal \(\beta^\ast=B u_j\), the risk is

$$ R(\hat\beta;B u_j) =\frac{B^2(u_j^Tv)^2}{(v^Tv)^2}\,v^T\Sigma v. $$

This follows directly by projecting \(B u_j\) onto \(v\), then evaluating the covariance-weighted squared length. The shared factor \(v^T\Sigma v/(v^Tv)^2\) is the same for every direction in this fixed design. The directional comparison is then controlled by \((u_j^Tv)^2\): how strongly the signal direction overlaps the nullspace. With a higher-dimensional nullspace, the overlap becomes a vector of components and each is weighted by its covariance eigenvalue.

A common mistake is to argue that a signal aligned with a high-variance direction must have greater risk because prediction errors in that direction are more costly. That reasoning considers only the weighting by \(\Sigma\) and ignores whether the interpolator makes an error in that direction, and how large that error is. The correction is to calculate \(P_{\mathcal{N}(X)}\beta^\ast\) first and then evaluate its covariance-weighted squared length.

These calculations are conditional on the observed design. If the design is random, its own distribution changes the nullspace from sample to sample, and an unconditional risk comparison requires averaging over that randomness. The fixed-design examples establish a precise geometric point, not a universal ranking for all random-design models: equal signal norms do not guarantee equal risks, and high- versus low-variance alignment alone does not determine the ordering.

Check Your Understanding

Use the nullspace projection formula to answer each question.

  1. Why does the minimum-norm interpolator’s coefficient error lie in the training nullspace?
  2. For a fixed design, what two ingredients determine the prediction risk of a signal?
  3. In the two-dimensional example, verify that the row vector \((1,2)\) is orthogonal to \((-2,1)\).
  4. Can a high-variance eigendirection signal have lower risk than an equal-norm low-variance signal? Explain what determines the answer.
  5. Why is a fixed-design comparison not, by itself, a conclusion about risk averaged over random training designs?
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